ellipse

[pbialos]

I was hoping you could give me a hint on how to find the equations of the axis of the ellipse of equation 5x^2-6xy+5y^2-4x-4y-4=0. I think this is supposed to be an exercise about lagrange multipliers or something related to the gradient, but i really dont know. I am clueless, i dont know any property about ellipses in general.

Many Thanks, Paul.

[TD]

Unfortunately, I don't know the English terminology that well regarding analytic geometry.

There is a thing which we call "middle line", which is a polar (polar line) of a point at infinity with respect to the conic (here the ellipse).

Axes are a special case of these lines, two of those whose directions are perpendicular, so where m_1 = - \frac{1}{{m_2 }}, where m is a direction.

For a general conic ax^2 + 2b''xy + a'y + 2by + 2b'x + a'' = 0 those direction are the solutions of b''m^2 + \left( {a - a'} \right)m - b'' = 0.

In this case, you get m = 1\,\,\, \vee \,\,m = - 1.

Now, the equations of the axes are then:

\begin{array}{l}
F_x ^\prime \left( {x,y} \right) + m_1 \cdot F_y ^\prime \left( {x,y} \right) = 0 \\
F_x ^\prime \left( {x,y} \right) + m_2 \cdot F_y ^\prime \left( {x,y} \right) = 0 \\
\end{array}

Here, F_x ^\prime \left( {x,y} \right) mean the partial derivative of the function to the variable x (same for y).

I tried it and it seems to be working, can you get the equations now?